Magnetic force on capacitor I have this problem: I have a capacitor formed by two parallel square-

Matthew Hubbard

Matthew Hubbard

Answered question

2022-05-07

Magnetic force on capacitor
I have this problem: I have a capacitor formed by two parallel square-shaped plates, of side a, and with distance d between them. They're asking me out of other things, the magnetic force as a function of some things. I calculate de magnetic force by calculating the gradient of the magnetic energy, given by:
E = 1 2 V B H d V
I have calculated B and H, that form, during the charge of the capacitor, loops around the center proportional to r, being r the distance from the central axis. The problem is that I have to integrate that to a cubic volume: a 2 d, one of those variables is independent, but the other two are not, and this problems never involve complicated integrals, such that one:
0 a 0 a r d x d y
If the plates were circles it would be trivial, but being squares, I'm sure I'm missing something.

Answer & Explanation

priffEmipsy4i37m

priffEmipsy4i37m

Beginner2022-05-08Added 17 answers

You want to know
0 a 0 a r d x d y
with
r = x 2 + y 2
via substitution x = α a and y = β a, you can bring it into the form
a 0 1 0 1 α 2 + β 2 d α d β .
For the remaining intgral, Wolfram alpha is your friend. With some more work you can show that in the exact result is
0 a 0 a r d x d y = a 3 [ 2 + log ( 1 + 2 ) ] .

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